t^2-16t-1=0

Solution for t^2-16t-1=0 equation:

t^2-16t-1=0

a = 1; b = -16; c = -1;
Δ = b2-4ac
Δ = -162-4·1·(-1)
Δ = 260
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{260}=\sqrt{4*65}=\sqrt{4}*\sqrt{65}=2\sqrt{65}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-2\sqrt{65}}{2*1}=\frac{16-2\sqrt{65}}{2}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+2\sqrt{65}}{2*1}=\frac{16+2\sqrt{65}}{2}
$